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I am Lyosha [343]
3 years ago
12

a circus performer launches himself from a springboard with an initial velocity of 21 m/s at an angle of 75 toward a platform ha

nging 20 m above the ground, a horizontal distance of 15 m away from the springboard. does he land on the platform or fall back down to the ground?

Physics
1 answer:
kherson [118]3 years ago
8 0

Answer:

The circus performer falls back down to the ground

Explanation:

The question parameters are;

The initial velocity of the circus performer = 21 m/s

The angle in which the performer launches himself = 75° towards the platform

The height of the platform above the ground = 20 m

The horizontal distance of the platform from the springboard = 15 m

The vertical motion of the circus performer is given by the following projectile motion relation;

y = y₀ + v₀·sinθ₀t-1/2·g·t²

Where;

y = Height reached by the circus performer

y₀ = Initial height of the the circus performer (the springboard) = 0 m

v₀ = Initial velocity of the the circus performer = 21 m/s

θ₀ = The angle with which the circus performer launches himself = 75°

t = The time of ,light of the circus performer

g = The acceleration due to gravity

Therefore, when the height is 20 m, we have;

20 = 21*sin(75)*t - 1/2*9.81*t²

Which gives;

21*sin(75)*t - 1/2*9.81*t² - 20 = 0

Factorizing using a graphing calculator, gives;

t = 1.623 or t = 2.513

Therefore, the circus performer passes the 20 m mark twice, in his motion, where the first time is when he is on his way up while the second time is when he is on his way down

The horizontal motion of the circus performer is given by the following projectile motion relation;

x = x₀ + v₀*cos(θ₀)* t

Where;

x₀ = The initial position of the circus performer in relation to the final position = 0

Plugging in the value of t when y = 20, we get;

x = 21×cos(75)×1.623 = 8.82 m, which is less than the 15 m platform distance from the spring board

Checking the other time value, we have;

x = 21×cos(75)×2.513 = 13.66 m which is also less than the 15 m platform distance from the spring board

Therefore, the circus performer misses the platform and falls back down to the ground.

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<u />x_{cm} = \frac{m_{1}x_{1}   + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}} = \frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12} = \frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41} = \frac{81}{41} = 1.9756

<u>y-values:</u>

<u />y_{cm} = \frac{m_{1}y_{1}   + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}} = \frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12} = \frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41} = \frac{-90}{41} = -2.1951

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(1.9756, -2.1951)

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